Thomas Calculus 13th [Solutions]

1212 Chapter 16 Integrals and Vector Fields
i j k
2i
2j k
3
23. Let F 2 yi 3 zj xk F 3 i j 2 k ; n F n 2
C
x y z
2y 3z x
2y dx 3z dy x dz F dr F n d 2 d 2 d , where
region enclosed by C on the plane S: 2x 2y z 2
C
S S S
S
d is the area of the
i j k
24. F
0
x y z
x y z
p N M P N M
y z z x x y
25. Suppose F Mi Nj Pk exists such that F i j k xi yj zk.
Then
Likewise,
P N 2 P 2
x
N
x y z x x y x y
( ) 1.
M P
2
M
2
P
y z x y y z y x
Summing the calculated equations
( y ) 1 and
N M 2 N 2
z
M
z x y z z x z y
2
P
2
P
2
N
2
N
2
M
2
M
x y y x z x x z y z z y
( ) 1.
3 or 0 3
(assuming the second mixed partials are equal). This result is a contradiction, so there is no field F such that
curl F xi yj zk.
26. Yes: If F 0 , then the circulation of F around the boundary C of any oriented surface S in the domain of F
is zero. The reason is this: By Stokes theorem, circulation F dr F n d 0 n d 0.
C
S
S
27.
2 2 4 2 2
2
4 2 2 2 2
r x y r x y F r 4x x y i 4y x y j Mi Nj
4
n F n
M N
C C C
x y
R
r ds ds M dy N dx dx dy
2 2 2 2 2 2 2 2 2 2
4 x y 8x 4 x y 8y dA 16 x y dA 16 x dA 16 y dA
R R R R
16I
16 I .
y
x
28.
2 2 2 2 2 2 2 2
P N M P N y x M y x y x y x
y z z x x 2 2 2 2
2 2 y 2 2 2 2 2 2
x y x y x y x y
0, 0, 0, 0, , curl F k 0.
However,
2 2
dr
dt
2 2
x y 1 r (cos t) i (sin t) j ( sin t) i (cos t)
j
dr
dt
F ( sin t) i (cos t) j F sin t cos t 1 F d r 1 dt 2 which is not zero.
C
2
0
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